Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Let f : R $$ \to $$ R be a function defined as

$$f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right.$$

Then, f is

$$f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right.$$

Then, f is

A

continuous if a = 0 and b = 5

B

continuous if a = –5 and b = 10

C

continuous if a = 5 and b = 5

D

not continuous for any values of a and b

Checking

if f(x) is continuous at x = 1 :

f(1^{$$-$$}) = 5

f(1) = 5

f(1^{+}) = a + b

if f(x) is continuous at x = 1,

then

f(1^{$$-$$}) = f(1) = f(1^{+})

$$ \Rightarrow $$ 5 = 5 = a + b

$$ \therefore $$ a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3^{$$-$$}) = a + 3b

f(3) = b + 15

f(3^{+}) = b + 15

if f(x) = is continuous at x = 3

then,

f(3^{$$-$$}) = f(3) = f(3^{+})

$$ \Rightarrow $$ a + 3b = b + 15 = b + 15

$$ \Rightarrow $$ a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5^{$$-$$}) = b + 25

f(5) = 30

f(5^{+}) = 30

if f(x) is continuous at x = 5 then,

f(5^{$$-$$}) = f(5) = f(5^{+})

$$ \Rightarrow $$ b + 25 = 30 = 30

$$ \Rightarrow $$ b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$$ \Rightarrow $$ a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$$ \therefore $$ f is not continuous for any value of a and b.

if f(x) is continuous at x = 1 :

f(1

f(1) = 5

f(1

if f(x) is continuous at x = 1,

then

f(1

$$ \Rightarrow $$ 5 = 5 = a + b

$$ \therefore $$ a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3

f(3) = b + 15

f(3

if f(x) = is continuous at x = 3

then,

f(3

$$ \Rightarrow $$ a + 3b = b + 15 = b + 15

$$ \Rightarrow $$ a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5

f(5) = 30

f(5

if f(x) is continuous at x = 5 then,

f(5

$$ \Rightarrow $$ b + 25 = 30 = 30

$$ \Rightarrow $$ b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$$ \Rightarrow $$ a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$$ \therefore $$ f is not continuous for any value of a and b.

2

$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$

A

exists and equals $${1 \over {2\sqrt 2 }}$$

B

exists and equals $${1 \over {4\sqrt 2 }}$$

C

exists and equals $${1 \over {2\sqrt 2 (1 + \sqrt {2)} }}$$

D

does not exists

$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$

If you put y = 0 at $${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$ it is in $${0 \over 0}$$ form. So we can use L' Hospital's Rule.

= $$\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$$

= $$\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$$

= $${1 \over {4\sqrt 2 }}$$

If you put y = 0 at $${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$ it is in $${0 \over 0}$$ form. So we can use L' Hospital's Rule.

= $$\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$$

= $$\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$$

= $${1 \over {4\sqrt 2 }}$$

3

For each x$$ \in $$**R**, let [x] be the greatest integer less than or equal to x.

Then $$\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$ is equal to :

Then $$\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$ is equal to :

A

$$-$$ sin 1

B

1

C

sin 1

D

0

$$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$$

$$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$$

4

Let f be a differentiable function from

**R** to **R** such that $$\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$$

for all $$x,y \in $$**R**.

If $$f\left( 0 \right) = 1$$

then $$\int\limits_0^1 {{f^2}} \left( x \right)dx$$ is equal to :

for all $$x,y \in $$

If $$f\left( 0 \right) = 1$$

then $$\int\limits_0^1 {{f^2}} \left( x \right)dx$$ is equal to :

A

1

B

2

C

$${1 \over 2}$$

D

0

$$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$$

$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$

$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$$

$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$$ $$ \Rightarrow f'\left( x \right) = 0$$

$$ \Rightarrow f\left( x \right) = $$ constant

as $$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$$

$$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$$

$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$

$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$$

$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$$ $$ \Rightarrow f'\left( x \right) = 0$$

$$ \Rightarrow f\left( x \right) = $$ constant

as $$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$$

$$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$$

Number in Brackets after Paper Name Indicates No of Questions

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

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Complex Numbers *keyboard_arrow_right*

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Vector Algebra and 3D Geometry *keyboard_arrow_right*

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Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

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Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*